Timeline for Group actions in a homotopy category
Current License: CC BY-SA 3.0
7 events
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| Jun 6, 2014 at 18:50 | comment | added | Justin Noel | . whose composite is multiplication by $|G|$ and which factors through $\mathrm{Ho}(M)^G(X,Y)$. From here one can see that $\mathrm{Ho}(M^G)(X,Y)\cong\mathrm{Ho}(M)^G(X,Y)$ when $|G|$ acts invertibly. It appears the non-trivial part is seeing that $\mathrm{tr}$ descends to the homotopy category. | |
| Jun 6, 2014 at 18:41 | comment | added | Justin Noel | As often happens, behind a vanishing spectral sequence argument lies a more elementary argument. My more involved method is probably overkill in this case. Nonetheless, the argument indicates why it is true. For $X,Y\in M^G$ we have a map $$ \mathrm{tr}\colon M(X,Y)\rightarrow M^G(X,Y)$$ given by $f(-)\mapsto \sum_{g\in G} (g\cdot f)(-)=\sum_{g\in G} gf(g^{-1} -)$. I suspect that $\mathrm{tr}$ descends to homotopy categories, perhaps by an argument with right homotopies. If this is the case, then precomposing with the restriction functor gives a self map on mapping sets in $\mathrm{Ho}(M^G)$.. | |
| Jun 6, 2014 at 15:47 | comment | added | Mike Shulman | Thanks! So in particular, if $M$ is additive, then every object is a homotopy monoid, so the result should apply. It's interesting that in the special case of actions by a group with invertible order one can prove it without such an involved calculation. | |
| Jun 6, 2014 at 15:20 | history | edited | Justin Noel | CC BY-SA 3.0 |
More precisely and correctly answered the question.
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| Jun 6, 2014 at 15:13 | history | edited | Justin Noel | CC BY-SA 3.0 |
More precisely and correctly answered the question.
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| Jun 6, 2014 at 14:53 | history | edited | Justin Noel | CC BY-SA 3.0 |
deleted 1 character in body
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| Jun 6, 2014 at 14:48 | history | answered | Justin Noel | CC BY-SA 3.0 |